What is Thermal Stress ?
Whenever there is some change (increase or decrease) in temperature of a body, it causes the body expand or contract. When the body expands or contracts freely corresponding to the rise or fall of temperature, we observe that there is no stress induced in that body. Then, what is thermal stress!
When the free expansion or contraction (deformation) of the body is prevented, some stresses are induced in that body. This stress is called thermal stress. That’s mean, thermal stress is not the stress that is induced only because of the change in temperature, it is the stress that is induced by the prevention of deformation due to change in temperature. The strain related to thermal stress is called thermal strain.
What is Thermal Stress in Simple Bars
We can easily calculate the thermal stress and thermal strain from the following discussion-
- There will be an amount of deformation due to the change in temperature. Assuming the bar free to expand or contract, the deformation of the bar can be measured.
- Calculate the required force or load to recover the deformed bar to the original length.
- The load or force will try to stop the deformation of the bar. Calculate the stress and strain in the bar for this load.
Let us consider a bar subjected to an increase in temperature,
l= Original length of the bar
δt= Increase of temperature
α= Coefficient of linear expansion
The increase of length for the increase in temperature,
δl = l*α*δt
Let us assume that, the ending part of the bar is fixed to a rigid support. So the body will experience compressive strain. That is-
ε =δl/l = (l*α*δt)/l = α*δt
Stress σ = ε* E = α*δt* E
This calculation is only valid if the supports are rigid. But if the supports yield by an amount equal to δ, then there is a little change in the calculation. That is-
δl = l*α*δt – δ
and strain, ε =δl/= (l*α*δt – δ)/l = α*δt- δ/l
Stress σ = ε* E= (α*δt- δ/l)* E
The values of the coefficient of linear expansion, α of the mostly used materials are given below
|Serial no||Material||coefficient of linear expansion
(per degree celcius)
|1||Steel||11.5* 10^-6 to 13* 10^-6|
|2||Wrought iron, cast iron||11* 10^-6 to 12* 10^-6|
|3||Aluminium||23* 10^-6 to 24* 10^-6|
|4||Copper, bronze, brass||17* 10^-6 to 18* 10^-6|
An example related to thermal stress are given below-
Two parallel walls 8m apart are stayed together by a steel rod of 30 mm diameter passing through metal plates and nuts at each end. The nuts are tightened home when the rod is at a temperature of 120 degree celcius. Determine the stress in the rod when the temperature falls to 60 degree celcius if
- The ends do not yield
- The ends yield by 1.5mm
Take the value of E= 200 Gpa and α= 12* 10^-6
Length, l= 8m
coefficient of linear expansion, α= 12* 10^-6
Modulus of elasticity, E= 200 Gpa= 200* 10^9
Temperature decrease, δt= 120-60= 60
Amount of yields at end, δ= 1.5mm= 1.5* 10^-3
- The ends do not yield
The decrease of length for the decrease in temperature,
δl = l*α*δt = 8* 12* 10^-6* 60 = 5.76 * 10^-3
Strain, ε =δl/l= (5.76 * 10^-3)/ 8 = 7.2* 10^-4
Stress in the rod, σa = ε* E= 7.2* 10^-4* 200* 10^9= 144* 10^6= 144Mpa
- The ends yield by 1.5mm= .0015m
Stress in the rod, σb = (α*δt- δ/l)*E= (12* 10^-6* 60- .0015/8)* 200* 10^9= 106.5* 10^6 = 106.5 Mpa
Related Post- Fundamental Concept of Stress and Strain